Specific Charge (e m) of the Electron

5- Specific Charge (e/m) of the Electron Abstract: We can measure the ratio between the charge and mass e/m (specific charge) of the electron using a fine-beam tube filled with helium gas. We get to see the path of the electron as it strikes the helium molecules, giving them energy that gets later released as electromagnetic waves (light). Introduction: When a charged electron passes through an electric field it will experience a force accelerating it in direction or opposite to the direction of the field depending on the charge. The magnitude of the force is 𝐹⃗ = 𝑞! 𝐸&⃗ 𝑞! is the charge of the electron and 𝐸&⃗ is the electric field. If the electric field is constant, then the acceleration that the electron will experience is constant as well. The electric field 𝐸&⃗ resulting from a voltage difference ∆𝑉 ∆# separated by a distance 𝑑 is constant and is equal to 𝐸&⃗ = . This means that $ when the electron leaves the area of the electric fiend it will have a kinetic energy: 1 𝐾𝐸 = 𝑞! ∆𝑉 = 𝑚𝑣 % 2 If then the moving electron enters a uniform magnetic field, then it will experience a force perpendicular to both the magnetic field and its speed. If the speed and the magnetic field are perpendicular, the path the electron will take would be circular. This force is responsible for the circular motion, so we can say that: 𝑚𝑣 % &⃗ = 𝐹⃗ = 𝑞! 𝑣⃗ × 𝐵 𝑟 &⃗ If 𝑣⃗ and 𝐵 are perpendicular, then 𝑚𝑣 % &⃗ 𝐹 = 𝑞! |𝑣⃗|4𝐵 4 = 𝑟 So, 𝐹 is constant. To measure the ratio between charge and mass, we can rearrange and get: 𝑞! 𝑣 = 𝑚! 𝐵𝑟 In our experiment, we use a voltage difference to accelerate the electron. We can reach an expression for speed from the kinetic energy: 𝑣=5 2𝑞! 𝑉 𝑚 & If we substitute that into '! we get: ! We notice that we got &! '! 62𝑞! 𝑉 𝑞! 𝑚 = 𝑚! 𝐵𝑟 on both sides. We can simplify: 𝑞! 𝑞! √2𝑉 =$ . 𝑚! 𝑚! 𝐵𝑟 𝑞! √2𝑉 = $ 𝑚! 𝐵𝑟 We finally get: 𝑞! 2𝑉 = " " 𝑚! 𝐵 𝑟 The magnetic field, 𝐵, is a result of current passing through a set of coils known as Helmholtz-coils. The field they produce is considered to beR uniform in the area between the two sets of coils. The magnitude of the field is: " 𝐵= ) # +, 𝜇( 8*9 - 𝑁 ≡ the number of coils. 𝐼 ≡ the current through the coils. 𝜇( ≡ permeability of vacuum. This field is measured in units of Tesla (T). R We will also be using a fine beam tube filled with helium gas to a pressure of 10 mm Hg. This beam tube will be connected to a source of voltage difference and has an electron gun, which has a cathode heated to create a stream of electrons via thermionic emission. These electrons will bend due to the electric field, and some will strike the helium atoms emitting light. This is visible as a ring of light. We can control the radius of the rings by changing: 1- the current through the coils, thus changing the magnetic field. By increasing the current, the field increases, thus the radius decreases. 2- the voltage difference between the cathode and anode in the electron gun, thus changing the speed of the electrons. By increasing the voltage, speed increases, thus the radius increases. Procedure: Set the components as shown: Part1: 1- set the current in the Helmholtz coils to 1A. 2- set the voltage of the anode to 140V and measure the diameter of the ring. 3- Increase the voltage by 20V and measure the diameter. 4- Repeat until you reach 220V. 5- Calculate the radius from the diameter you measured, and 𝑟 % . 6- Plot 𝑉 − 𝑟 % and find the slope. ! 7- Using the slope, find '. Part2: 1- Set the voltage of the anode at 180V. 2- Set the current in the coils to 0.9A and measure the diameter of the ring. 3- Increase the current by 0.05A and measure the diameter. 4- Repeat until you reach 1.15A. 5- Calculate the radius from the diameter you measured, and 𝑟 % . 6- Calculate the field B from the current, then find 7- Plot 𝑟 " − 1 and find the slope. 𝐵2 # $! . ! 8- Using the slope, find '. Data: Part1: 𝑠𝑙𝑜𝑝𝑒 = / 𝑉 (V) 𝑟 (cm) 𝑟 % × 10.) (m) 140 4.47 19.98 160 4.80 23.04 180 5.10 26.01 200 5.39 29.05 220 5.66 32.03 ∆𝑉 152 − 56 = = 68.57 × 10/ 𝑉 ⁄𝑚% % .) (22 − 8) × 10 ∆𝑟 / 4 % 𝑁𝐼 4 % (120)(1) 𝐵 = 𝜇( L N = (1.25 × 10.0 ) L N = 7.1554 × 10.) 𝑇 5 𝑅 5 0.15 𝑒 2𝑉 2𝑠𝑙𝑜𝑝𝑒 2(68.57 × 10% ) = " "= = = 2.679 × 10## 𝐶 ⁄𝑘𝑔 " &' " (7.1554 × 10 ) 𝑚 𝐵 𝑟 𝐵 Part2: 1 1 - / (× 10# ) 𝐵" 𝑇 𝑟 (cm) 𝑟 % (× 10.) ) (m) 4.4721 5.000 5.75 33.0625 0.95 4.7206 4.4875 5.48 30.0304 1.00 4.9690 4.0500 5.20 27.0400 1.05 5.2175 3.6735 4.90 24.0100 1.10 5.4659 3.3471 4.58 20.9764 1.25 5.7144 3.0624 4.24 17.9776 𝐼 (A) 𝐵 (T) (× 10.) ) 0.90 𝑠𝑙𝑜𝑝𝑒 = ∆𝑟 % 1% ∆ 𝐵 = ∆𝑟 % ∆𝐵 % 31.8 × 10.) − 0 = = 7.7561 × 10.12 𝑚% 𝑇 % (4.75 − 0.6) × 100 𝑒 2𝑉 2𝑉 2(180) ## ⁄ = " "= = 𝐶 𝑘𝑔 −10 = 4.6415 × 10 𝑚 𝐵 𝑟 𝑠𝑙𝑜𝑝𝑒 7.7561 × 10 Sources of error: 1- Accuracy in measurement tools. 2- Human error. Not reading the measurement accurately. 3- Approximation. 4- Slope error. Error: Part1: 2.679 × 10"" − 1.7588 × 10"" 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = × 100% 1.7588 × 10"" = 52.3% Part2: 4.6415 × 10"" − 1.7588 × 10"" 𝑝𝑒𝑟𝑐𝑒𝑛𝑡𝑎𝑔𝑒 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = × 100% 1.7588 × 10"" = 163.9% Part 1: 𝑉 𝑟 % × 10.) 𝑒⁄ (× 1011 ) 𝑚 ⁄𝑚[ Z𝑒⁄𝑚 − 𝑒SSSSSS 11 × 10 𝑒⁄ [ Z𝑒⁄𝑚 − SSSSSS 𝑚 140 19.98 2.737 0.032 1.024× 1016 160 23.04 2.713 0.008 6.4× 1017 180 26.01 2.703 -0.002 4× 1010 200 29.05 2.687 -0.018 3.24× 1018 220 32.03 2.683 -0.022 4.84× 1018 % SSSSSS 𝑒⁄ = 2.705 × 1011 𝑚 2 𝑒⁄ − 𝑒SSSSSS ∑ ⁄ ( ) 𝑚 𝑚 ∆ 𝑒⁄𝑚 = 5 = 974.7 × 100 𝐶Y𝑘𝑔 𝑁(𝑁 − 1) Part 2: 𝑟 % × 10.) 𝑒⁄ (× 1011 ) 𝑚 𝑒⁄ [ Z𝑒⁄𝑚 − SSSSSS 𝑚 × 1011 ⁄𝑚[ Z𝑒⁄𝑚 − 𝑒SSSSSS 5.000 33.0625 5.44 -0.16 2.56 × 10%2 4.4875 30.0304 5.38 -0.22 4.84× 10%2 4.0500 27.0400 5.39 -0.21 4.41× 10%2 3.6735 24.0100 5.50 -0.1 1 × 10%2 3.3471 20.9764 5.75 0.15 2.25 × 10%2 3.0624 17.9776 6.13 0.53 2.81 × 10%1 # $! % SSSSSS 𝑒⁄ = 5.598 × 1011 𝑚 𝑒⁄ ∑(𝑒⁄𝑚 − SSSSSS 𝑚 ∆ 𝑒⁄𝑚 = 5 𝑁(𝑁 − 1) 2 ) = 1.199 × 1012 𝐶Y𝑘𝑔 Discussion and Conclusion: We were able to find the ratio between the charge and mass of the electron using the fine tube. Electrons experience a centripetal acceleration due to the magnetic perpendicular to its velocity and collide with the helium atoms that emit light. 1- Electrons accelerate in electric fields. 2- Electrons gain centripetal acceleration due to the magnetic field. 3- If the magnetic field and the speed are perpendicular, the path of the electrons will be circular. 4- If the magnetic field and the speed are parallel, no magnetic force will be exerted. 5- If the electron enters the area of the magnetic field at an angle, the path will be spiral. 6- The magnetic field resulting from Helmholtz coils is directly proportional to the number of coils and the current, and inversely proportional to the radius. 7- The magnetic force is directly proportional to the speed of the electron and the magnitude of the field. It Is also dependent on the angle between the speed and the field. 8- We can approximate the value of e/m experimentally by: - Changing the voltage that the electrons accelerate at. - Changing the magnetic field affect the electrons by changing the current in the coils. 9- Increasing the current (and thus the magnetic field) will decrease the radius of the path while increasing the voltage (and thus the speed of the electron) will increase the radius of the path. 10- The value of ") $! * ! is always constant and equal to ! ' = 1.7588 × 1011 .